One mole of an ideal monoatomic gas is taken round cyclic process $A B C A$ as shown in figure. Calculate the maximum temperature attained by the gas during the cycle?

3 T_{1}

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2 T_{1}

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4 T_{1}

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5 T_{1}

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Solution

The correct option is C $4 T_{1}$
Maximum temperature will lie between $B$ and $C$ .
$P V$ equation for the product $B C$ can be written as
$P = - m V + C$
here $m = \frac{2 P_{o}}{2 V_{o}} = \frac{P_{o}}{V_{o}} a n d C \Rightarrow \frac{y - 3 P_{o}}{V_{o} - 0} \frac{P_{o}}{V_{o}} =\Rightarrow y = 4 P_{o} C = 4 P_{o}$
$P = - (\frac{P_{o}}{V_{o}}) V + {4 P}_{o} (1) P V = - (\frac{P_{o}}{V_{o}}) V^{2} + {4 P}_{o} V R T = - (\frac{P_{o}}{V_{o}}) V^{2} + {4 P}_{o} V [n = 1 \Rightarrow P V = R T] T = \frac{1}{R} [{4 P}_{o} V - (\frac{P_{o}}{V_{o}}) V^{2}] (2)$
For $T$ to be maximum, $\frac{d T}{d V} = 0$
${4 P}_{o} - 2 \frac{P_{o}}{V_{o}} V = 0 {4 P}_{o} = 2 \frac{P_{o}}{V_{o}} V V = 2 V_{o}$
Substituting it in equation (2), we get
$T_{m a x} = \frac{1}{R} [{4 P}_{o} (2 V_{o}) - (\frac{P_{o}}{V_{o}}) {(2 V_{o})}^{2}] = \frac{1}{R} [{8 P}_{o} V_{o} - {4 P}_{o} V_{o}] T_{m a x} = \frac{{4 P}_{o} V_{o}}{R} = 4 T_{1}$

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